Question

$\displaystyle\int \frac{2\,dx}{(e^{x} + e^{-x})^{2}}$ is equal to

• A. $\dfrac{e^{2x}}{1+e^{2x}} + c$
• B. $\dfrac{e^{-2x}}{1+e^{-2x}} + c$
• C. $\dfrac{1}{1+e^{2x}} + c$
• D. $-\dfrac{1}{1+e^{2x}} + c$

Hint:

$(e^x+e^{-x})^2 = e^{-2x}(1+e^{2x})^2$. Multiply numerator and denominator by $e^{2x}$ to set up a standard substitution.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
Rewrite $(e^x+e^{-x})^2$ in terms of $e^{2x}$ and then use a substitution.
Step 2: Detailed Explanation:
$(e^x+e^{-x})^2 = e^{-2x}(e^{2x}+1)^2$.
So $\displaystyle\int\frac{2\,dx}{(e^x+e^{-x})^2} = \int\frac{2e^{2x}}{(e^{2x}+1)^2}\,dx$.
Let $u = e^{2x}+1$, $du = 2e^{2x}dx$: integral $= \displaystyle\int\frac{du}{u^2} = -\dfrac{1}{u}+c = -\dfrac{1}{1+e^{2x}}+c$.
Step 3: Final Answer:
$-\dfrac{1}{1+e^{2x}} + c$.